Consider an integrable function $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $af(a)+bf(b)=0$ when $ab=1$. Find the value of the following integration:
$$ \int_{0}^{\infty} f(x) \,dx $$
Hasin_Ahmad wrote:
Consider an integrable function $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $af(a)+bf(b)=0$ when $ab=1$. Find the value of the following integration:
$$ \int_{0}^{\infty} f(x) \,dx $$
$xf(x)+\frac 1xf(\frac 1x)=0$ and so $f(\frac 1x)=-x^2f(x)$
Setting $x=\frac 1t$ in $\int_0^{+\infty}f(x)dx$, we get $\int_0^{+\infty}f(x)dx$ $=\int_{+\infty}^0f(\frac 1t)\frac{-dt}{t^2}=\int_{+\infty}^0f(t)dt$ $=-\int_0^{+\infty}f(t)dt$
And so $\boxed{\int_0^{+\infty}f(x)dx=0}$